Question

Find the interval in which f(x) = ∫2√2sin²t + (2 – √2)sint – 1)dt for t ∈ [0, x] where 0 < x < 2π is increasing or decreasing.

Answer 1

We have

⇒ f'(x) = (2(√2 sin²x + (2 – √2)sinx – 1)

= (2sinx – 1)(√2sinx + 1)

When f(x) is increasing, so f'(x) > 0

⇒ (2 sin – 1)(√2 sinx + 1) > 0

⇒ (√2 sinx + 1) < 0, (2sinx – 1) > 0

⇒ sinx < – 1/√2 , sinx > 1/2

⇒ x ∈ (5π/4 , 7π/4) and x ∈ (π/6 , 7π/6)

⇒ x ∈ (5π/4 , 7π/4) ∪ (π/6 , 7π/6)

when f(x) is decreasing, so

⇒ f'(x) < 0

⇒ (2sinx – 1)(√2sinx + 1) < 0

⇒  –1/√2 < sinx < 1/2

⇒ x ∈ (0, π/6) ∪ (5π/6 , 5π/4 ) ∪ (7π/4 , 2π)